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### Rational Inequality

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Date: 10/09/2001 at 19:57:42
From: alan
Subject: Algebra, rational inequality

I have a question about how to solve a rational inequality and give
an answer in interval notation. The example looks like this.

- ____5____  greater than or equal to _5_
3h + 2                             h

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Date: 10/10/2001 at 18:12:13
From: Doctor Greenie
Subject: Re: Algebra, rational inequality

Hi, Alan -

Let's find the solution first by looking at the graphs of the two
rational expressions separately and analyzing them. Later we can look
at a purely algebraic way to reach the same solution.

I. Solution by analyzing graphs

(1) -5/(3h+2)

At h = -2/3, the denominator is zero and the value is undefined. For
values of h larger than -2/3, the denominator is positive, so the
function value is negative; for values of h less than -2/3, the
denominator is negative, so the function value is positive. For large
negative values of h, the function value is a small positive number;
for large positive values of h, the function value is a small negative
number.

In summary, the value for this function is

- small positive for large negative values of h;
- increasingly large positive as you move towards h=-2/3, becoming
very large positive for values of h slightly less than -2/3;
- undefined for h=-2/3;
- very large negative for values of h slightly larger than -2/3;
- increasing (less negative) as you move towards values of h
approaching infinity; small negative for large positive values
of h

(2) 5/h

At h = 0, the denominator is zero and the value is undefined. For
values of h larger than 0, the denominator is positive, so the
function value is positive; for values of h less than 0, the
denominator is negative, so the function value is negative. For large
negative values of h, the function value is a small negative number;
for large positive values of h, the function value is a small positive
number.

In summary, the value for this function is

- small negative for large negative values of h;
- increasingly large negative as you move towards h=0, becoming
very large negative for values of h slightly less than 0;
- undefined for h=0;
- very large positive for values of h slightly larger than 0;
- decreasing (less positive) as you move towards values of h
approaching infinity; small positive for large positive values
of h

With the given inequality, we want to find the values of h for which
the value of function (1) is greater than or equal to the value of
function (2).

(a) Function (1) is positive for all values of h less than -2/3;
function (2) is negative for all negative values of h. Therefore, the
inequality is satisfied for all values of h less than -2/3.

(b) Function (1) is negative for all values of h greater than -2/3;
function (2) is positive for all positive values of h. Therefore, the
inequality is not satisfied for any values of h greater than 0.

(c) In the interval between h-values of -2/3 and 0, both functions are
negative; the function (1) value is rapidly increasing from negative
infinity towards 0; the function (2) value is decreasing rapidly
toward negative infinity. Somewhere in the interval the two function
values will be the same; the inequality will be satisfied in the
interval between that value and 0.

To find the value where the two function values are equal, we solve
the equation associated with the given inequality:

-5    5
---- = -
3h+2   h

Multiplying both sides of the equation by the two denominators to
clear fractions, we have

-5h = 5(3h+2)
-5h = 15h+10
-20h = 10
h = -1/2

We have determined, by analyzing the graphs of the two functions,
that the inequality is satisfied for values of h which are either in
(-infinity,-2/3) or in [-1/2,0). The two functions have equal values
at h = -1/2, so equality holds there and so h=-1/2 is included in the
solution set of the "greater than or equal to" inequality, as
indicated by the square bracket. One or the other of the functions is
undefined at h=0 and h = -2/3, so those points are not included in the
solution set of the inequality, as indicated by the curved brackets.

II.  Algebraic solution

Now let's look at a couple of different ways to arrive at the same
solution set algebraically, without analyzing the graphs of the
functions.

Probably the easiest way to find the solution set to the inequality

-5     5
---- >= -
3h+2    h

is to rewrite the inequality as

-5    5
---- - - >= 0
3h+2   h

Then we can combine the fractions on the left:

-5(h)-5(3h+2)   -5h-15h-10   -10(2h+1)
------------- = ---------- = ---------
h(3h+2)        h(3h+2)     h(3h+2)

This rational function is undefined (denominator is 0) at h=0 and
h=-2/3; it has the value 0 (numerator is 0) when h=-1/2. These three
critical points divide the set of possible h-values into four
intervals; we need to examine the function in each of those intervals
to see whether the function value is positive or negative, as
determined by the signs of the two factors in the numerator and the
two factors in the denominator. The original inequality is satisfied
when this function value is positive. We have

(-)(-)
(-infinity,-2/3):  ------ = (+)  YES
(-)(-)

(-)(-)
(-2/3,-1/2):       ------ = (-)  NO
(-)(+)

(-)(+)
(-1/2,0):          ------ = (+)  YES
(-)(-)

(-)(+)
(0, infinity):     ------ = (-)  NO
(+)(+)

This analysis shows the strict inequality is satisfied on the
intervals (-infinity, -2/3) and (-1/2,0); and we know the equality
is satisfied at h=-1/2. So, as before, our solution set is
(-infinity, -2/3) and [-1/2,0).

A less common algebraic approach to the solution would be to multiply
the original inequality on both sides by the two denominators. When
you do this, you need to remember that if you multiply both sides of
an inequality by a negative number, the direction of the inequality is
reversed.

-5     5
---- >= -
3h+2    h

and we multiply both sides by (h)(3h+2). The factor (h) is negative
for h-values less than 0 and positive for h-values greater than 0; the
factor (3h+2) is negative for h-values less than -2/3 and positive for
h-values greater than -2/3. The product (h)(3h+2) is therefore

(a) positive (negative * negative) for h-values less than -2/3; or
(b) negative (negative * positive) for h-values between -2/3 and 0; or
(c) positive (positive * positive) for h-values greater than 0

After multiplying the inequality by (h)(3h+2) and reversing the
direction of the inequality when needed, we have

(a)  -5h >= 5(3h+2) and h < -2/3
-5h >= 15h+10 and h < -2/3
-20h >= 10 and h < -2/3
h <= -1/2 and h < -2/3
h < -2/3

or

(b)  -5h <= 5(3h+2) and -2/3 < h < 0
-5h <= 15h+10 and -2/3 < h < 0
-20h <= 10 and -2/3 < h < 0
h >= -1/2 and -2/3 < h < 0
-1/2 <= h < 0

or

(c)  -5h >= 5(3h+2) and h > 0
-5h >= 15h+10 and h > 0
-20h >= 10 and h > 0
h <= -1/2 and h > 0
(no solutions here...)

And so our solution set, from parts (a) and (b), is once again
(-infinity,-2/3) or [-1/2,0).

I hope all this helps and that the several different methods do not
confuse you. I find I am much more comfortable and confident with a
problem where I know I have several different methods I can try to use
in case I have trouble using one of the methods; or if an answer I get
by one method doesn't seem right and I want to check it by using
another method.

Write back if you have any further questions on this topic.

- Doctor Greenie, The Math Forum
http://mathforum.org/dr.math/
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Associated Topics:
High School Basic Algebra
High School Equations, Graphs, Translations
High School Functions

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